Precalculus Examples

$\frac{3 x (x - 1)}{2 x^{2} - 5 x - 3}$

Step 1

Set the denominator in $\frac{3 x (x - 1)}{2 x^{2} - 5 x - 3}$ equal to $0$ to find where the expression is undefined.

$2 x^{2} - 5 x - 3 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor by grouping.

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Step 2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = - 5$ .

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Step 2.1.1.1

Factor $- 5$ out of $- 5 x$ .

$2 x^{2} - 5 x - 3 = 0$

Step 2.1.1.2

Rewrite $- 5$ as $1$ plus $- 6$

$2 x^{2} + (1 - 6) x - 3 = 0$

Step 2.1.1.3

Apply the distributive property.

$2 x^{2} + 1 x - 6 x - 3 = 0$

Step 2.1.1.4

Multiply $x$ by $1$ .

$2 x^{2} + x - 6 x - 3 = 0$

Step 2.1.2

Factor out the greatest common factor from each group.

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Step 2.1.2.1

Group the first two terms and the last two terms.

$(2 x^{2} + x) - 6 x - 3 = 0$

Step 2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$x (2 x + 1) - 3 (2 x + 1) = 0$

Step 2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$(2 x + 1) (x - 3) = 0$

Step 2.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 x + 1 = 0$

$x - 3 = 0$

Step 2.3

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Step 2.3.2

Solve $2 x + 1 = 0$ for $x$ .

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Step 2.3.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 2.3.2.2

Divide each term in $2 x = - 1$ by $2$ and simplify.

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Step 2.3.2.2.1

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Step 2.3.2.2.2

Simplify the left side.

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Step 2.3.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Step 2.3.2.2.3

Simplify the right side.

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Step 2.3.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 2.4

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 2.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.5

The final solution is all the values that make $(2 x + 1) (x - 3) = 0$ true.

$x = - \frac{1}{2}, 3$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq - \frac{1}{2}, 3}$

Step 4